# Voltage divider

A very common and useful resistor circuit goes by the nickname *voltage divider*. We will work out how this circuit works, and we’ll see where the nickname comes from.

The voltage divider is really quite simple, and you could analyze it from scratch every time you come across one. But it is *so* common it is a good idea to prepare ahead of time so you can pop out the answer on the spot.

Written by Willy McAllister.

### Contents

- What is a voltage divider?
- The voltage divider equation
- Voltage divider practice problems
- What’s in a nickname?

### Where we’re headed

A “voltage divider” is the pattern of two resistor in series, with an input voltage applied to both ends and an output voltage measured across one of the resistors.

The output voltage is a function of the input voltage and the ratio of resistor values,

$v_{out} = v_{in}\,\dfrac{\text R2}{\text R1 + \text R2}$

## What is a voltage divider?

A voltage divider looks like this,

It is a simple circuit with two resistors in series. One voltage is connected to the top and bottom, and another voltage is measured across one of the resistors. We think of the voltage across the top and bottom as the input voltage, while the voltage across the single resistor is the output.

The little circles indicate the *ports* of the voltage divider, and they are connected to something we can’t see right now.

## The voltage divider equation

Our goal is to come up with an expression that relates output $v_{out}$ to input $v_{in}$. The answer will resemble this,

$v_{out} = v_{in}\times(\text{something})$

This is a very simple series circuit so it won’t take a lot of effort to figure out.

Before we start, we make this important *Assumption*,

**Assume the current flowing out of the divider from its center node is zero.**

This is a good place to pause. You understand Ohm’s Law and series resistors. See if you can come up with an expression for $v_{out}$ in terms of $v_{in}$ on your own.

… … Okay great. You’ve derived a voltage divider expression. Now I will have a try.

A good place to start is to find the current through $\text{R1}$ and $\text{R2}$.

Here’s the one tricky step in analyzing a voltage divider: Based on the assumption, we know $\text{R1}$ and $\text{R2}$ have the same current, so we can consider them to be in series.

$i_{\text{1}} = i_{\text{2}}\qquad$ and for now let’s just call this $i$.

Let’s write an equation using Ohm’s Law, $v = i\,\text{R}$, and what we know about resistors in series: resistors in series add.

$v_{in} = i \,(\text R1 + \text R2)$

Rearrange the equation to solve for $i$,

$i = v_{in}\,\dfrac{1}{\text R1 + \text R2}$

This tells us current $i$ in terms of $v_{in}$ and both resistors.

Next, we write an expression for $v_{out}$ using Ohm’s Law,

$v_{out} = i \, \text R2$

$i = \dfrac{v_{out}}{\text R2}$

We substitute for $i$ in the previous equation to get,

$\dfrac{v_{out}}{\text R2} = v_{in}\,\dfrac{1}{\text R1 + \text R2}$

and, after moving $\text R2$ over to the right side, we derive the voltage divider equation,

$v_{out} = v_{in}\,\dfrac{\text R2}{\text R1 + \text R2}$

### Resistor ratio

The resistor ratio term is always less than $1$. Resistance is always a positive number. So this is true for any values of $\text{R1}$ and $\text{R2}$. (Take a second to convince yourself of this.)

Since the resistor ratio is less than $1$, that means $v_{out}$ is always less than $v_{in}$. Input voltage $v_{in}$ is divided down to $v_{out}$ by a fixed ratio determined by the resistor values. This is where the circuit gets its nickname: *voltage divider*.

### Example

**Find $v_{out}$ for this voltage divider**,

We insert the input voltage and resistor values into the voltage divider equation. Remember, the *bottom* resistor, $\text{R2}$, appears in the numerator.

$v_{out} = v_{in}\,\dfrac{\text R2}{\text R1 + \text R2}$

$v_{out} = 12\,\text V\cdot\dfrac{3\,\text k\Omega}{1\,\text k\Omega + 3\,\text k\Omega}$

$v_{out} = 12\,\text V\cdot\dfrac{3\,\text k\Omega}{4\,\text k\Omega}$

$v_{out} = 12\,\text V\cdot\dfrac{3}{4} = 9 \,\text V$

Open this simulation model and click on **DC** in the menu to confirm the DC operating point.

We finish up with two optional steps,

**Find the current and power.**

Use Ohm’s Law to find the current,

$i = \dfrac{v_{in}}{\text R1 + \text R2} = \dfrac{12\,\text V}{1\,\text k\Omega + 3\,\text k\Omega} = \dfrac{12\,\text V}{4\,\text k\Omega} = 3\,\text{mA}$

Knowing the current, we compute the power dissipated by our voltage divider,

$p = i\,v = 3\,\text{mA} \cdot 12\,\text V = 36\,\text{mW}$

Summary: Our voltage divider takes an input voltage (in this case $12\,\text V$, but it could be any value) and divides it down to create an output voltage $v_{out}$ that’s $3/4$ of its input voltage. The $3/4$ ratio is determined by our choice of the two resistor values. A $3\,\text{mA}$ current flows through the voltage divider, causing it to dissipate $36\,\text{mW}$.

## Voltage divider practice problems

Do these practice problems two ways. First work out the answer analytically with pencil and paper. Then simulate your answer using the Circuit Sandbox.

### Problem 1

Let $v_{in}= 6\,\text V$, $\text R1=50\,\text k\Omega$, and $\text R2=10\,\text k\Omega$.

**Find $v_{out}$**

$v_{out} =$ _________ $\,\text V$

## show answer

$v_{out} = v_{in}\,\dfrac{\text{R2}}{\text{R1} + \text{R2}}$

$v_{out}= 6\,\text V \cdot \dfrac{10\,\text{k}\Omega}{10\,\text{k}\Omega+50\,\text{k}\Omega}$

$v_{out}= 6\,\text V \cdot \dfrac{10\,\text{k}\Omega}{60\,\text{k}\Omega} = 6\,\text V \cdot \dfrac{1}{6}$

$v_{out} = 1\,\text V$

Simulation model of Problem 1. Click on **DC** in the top menu to find the operating point.

### Problem 2

Let $\text R1=90\,\text k\Omega$, $\text R2=10\,\text k\Omega$, and $v_{out}= 1.5\,\text V$.

**Find $v_{in}$.**

$v_{in} =$ _________ $\text V$

## show answer

$v_{out} = v_{in}\,\dfrac{\text{R2}}{\text{R1} + \text{R2}}$

$1.5\,\text V= v_{in}\, \dfrac{10\,\text{k}\Omega}{10\,\text{k}\Omega+90\,\text{k}\Omega}$

$1.5\,\text V= v_{in}\, \dfrac{1}{10}$

$v_{in} = 10 \cdot 1.5\,\text V$

$v_{in} = 15\,\text V$

Simulation model of Problem 2. Double-click on the voltage source and enter a voltage value for $v_{in}$ to get the output voltage you want. Repeat the **DC** analysis to confirm your choice.

### Problem 3

Let $v_{in}= 5\,\text V$, $v_{out}=2\,\text V$, and $\text R1=30\,\text k\Omega$.

**Find $\text R2$.**

$\text R2 =$ _________ $\Omega$

## show answer

$v_{out} = v_{in}\,\dfrac{\text{R2}}{\text{R1} + \text{R2}}$

$2\,\text V = 5 \,\text V \cdot \dfrac{\text{R2}}{30\,\text k\Omega + \text{R2}}$

$30\,\text k\Omega + \text R2 = \dfrac{5\,\text V}{2\,\text V} \, \text R2$

$30\,\text k\Omega = \left (\dfrac{5}{2} \,\text R2 \right ) - \text R2 = \dfrac{3}{2} \,\text R2$

$\text R2 = \dfrac{2}{3}\,30\,\text k\Omega$

$\text R2 = 20\,\text k\Omega = 20000 \,\Omega$

Check by plugging $\text R2$ back into the voltage divider equation:

$v_{out} = 5 \,\text V \cdot \dfrac{20\,\text k\Omega}{30\,\text k\Omega +20\,\text k\Omega} = 5 \cdot \dfrac{20}{50} = 2\,\text V \qquad \checkmark$

Simulation model of Problem 3. Double-click on $\text R2$ and enter a resistance value to get the desired output voltage. Repeat the **DC** analysis to confirm your choice.

### Problem 4 - design challenge

Let $v_{in}= 1\,\text V$, $v_{out}=\dfrac{v_{in}}{2}$.

**Design a voltage divider that dissipates $10 \,\mu\text{W}$.**

$\text R1 =$ _________ $\Omega\qquad$

$\text R2 =$ _________ $\Omega$

## show answer

$v_{out} = v_{in}\,\dfrac{\text{R2}}{\text{R1} + \text{R2}}$

Let's start by figuring out what the relationship has to be between $\text R1$ and $\text R2$ to make $v_{out} = v_{in}/2$.

We'll do this symbolically,

$\dfrac{1}{2} v_{in} = v_{in}\,\dfrac{\text{R2}}{\text{R1} + \text{R2}}$

$\dfrac{1}{2} = \dfrac{\text{R2}}{\text{R1} + \text{R2}}$

$\text{R1} + \text{R2} = 2\,\text R2$

$\text R1 = 2\,\text R2 - \text R2$

$\text R1 = \text R2$

The two resistors have the same value. We don't know what the value is, yet. Use the power constraint to discover the allowed current through the divider. Tip: When you calculate the power in a voltage divider, be sure to count the power dissipated by *both* resistors, not just the bottom one.

$p = i \cdot v\qquad$ formula for power in a resistor

$10\,\mu\text W = i \cdot v_{in} = i \cdot 1\,\text V$

$i = \dfrac{10\,\mu\text W}{1\,\text V} $

$i = 10 \,\mu\text A$

Given this current, plus the input voltage supplied by the problem statement, the top-to-bottom resistance of the divider must be,

$\text R1+\text R2 = \dfrac{v}{i}$

$\text R1+\text R2 = \dfrac{1\,\text V}{10\,\mu\text A} = \dfrac{1}{10 \times 10^{-6}} = 1 \times 10^5 = 100\,\text k\Omega$

We know the resistors are the same value, and now we know their sum, so,

$\text R1 = \text R2 = 50 \,\text k\Omega$

Simulation model of Problem 4. Double-click on both resistors and give them resistance values to achieve the design specifications: $v_{out} = v_{in}/2$ and $p = 10 \,\mu\text{W}$.

## What’s in a nickname?

We mentioned the nickname of this circuit is a *voltage divider*. In many situations, that is exactly what it does. However, remember back at the beginning we made an assumption, that the current leaving the divider is zero, or very close to zero? Under certain conditions (that we’ll cover in the next article), the actual output voltage might be slightly lower than the value predicted by the voltage divider equation. The lesson: Call the circuit by its nickname, but remember, it’s *only* a nickname.

## Summary

We give the name “voltage divider” to a pattern of resistors that looks like this, with an input voltage applied to both ends, and an output voltage measured across one of the resistors.

The output voltage is a function of the input voltage and a ratio of resistor values,

$v_{out} = v_{in}\,\dfrac{\text{R2}}{\text{R1} + \text{R2}}$

where $\text{R2}$ is the resistor on the bottom of the divider.

The next article covers some of the design subtleties of voltage dividers such as real-world resistor tolerance and non-zero output current.

## Questions

Can you look at this circuit? https://photos.app.goo.gl/rvxFGtpx8f0ohlS03 Using voltage divider law, does that make the voltage at the 555 chip #7 input equal to +4.08V ?

Does C1 have any effect on the input at 7?

Also look at this circuit: https://photos.app.goo.gl/M1UIqOvtjKOBKdxy1 and resistor values here: https://photos.app.goo.gl/BPD1LuFjnpNpHRWc2

Using voltage divider laws again it appears voltage at 555 input #5 is +3V. However, according to this information on the 555 timer chip it takes a supply voltage of +4.5V to +16V. 3 is under the minimum of range…whats going on here, or did I do the calculation wrong??? https://photos.app.goo.gl/DKiRKaET2OkI6Rx63

Thank you for doing this site and the KA videos!

The 555 is not a simple chip to understand. Your circuit is an astable multivibrator (http://www.ti.com/lit/ds/symlink/lm555.pdf, page 10, 7.4.2). R1 and R2 are configured as a voltage divider, but there’s a lot of action happening at the two nodes on either side of R2. The voltage divider assumption (negligible current flowing out of the middle node of the divider) is not met for this application. Question 1: The voltage at pin 7 is not a stable value, it moves up and down, as controlled by the 555. You should see a triangle wave at pin 7. Here’s why…

How it works: The capacitor charges up through R1+R2. Voltage rises as you would expect an RC circuit to do. The 555 keeps an eye on Vcap with Thresh (pin 6) and Trig (pin2). When Vcap gets up to 2/3 of VCC (6v * 2/3 = 4v) the Thresh pin says Ha! and turns on Discharge pin 7. The Discharge pin pulls current out of the capacitor through R2, and Vcap reverses direction and goes back down. When Vcap gets down to 1/3 of VCC (6 * 1/3 = 2v), that gets sensed by the Trig pin, #2. Inside the 555, the Trig says Ha! and causes the Discharge pin to turn off. Current to start collecting on the capacitor again and Vcap rises, just like it did at the beginning. Question 2: The capacitor has a lot of effect on the voltage at pin 7.

Question 3: See the schematic on page 3 of the TI data sheet. Notice the voltage divider inside the 555 made with 3 identical R’s (5 kohms each). When you connect your 10K external resistors to pin 5 you are hooking them up in parallel with the internal divider. That interferes with the simple voltage divider expression you used to get 6v * 10k/(10k + 10k). Pin 5 is connected to one of the middle nodes if the internal voltage divider. It’s not connected directly to VCC. Pin 5 does not have to obey the same min/max rule that applies to VCC.

Wow, I have a long way to go in learning here! I haven’t got to capacitors yet here on your site but have a “65 in 1” kit that had this 555 circuit as the first project and I thought I recognized and understood the resistor divider section but not yet… Thank you

I suggest you build the sample project by rote, just doing what the instructions say. You can leave Pin 5 unconnected to start with. If you can get your circuit to blink the LED, then you have your foot in the door. In this circuit, the capacitor acts like a bucket for charge. If you provide a current to the capacitor, charge builds up and the voltage rises. If you take away charge, the voltage drops. If you multiply R * C you will get the “time constant” of the circuit, measured in seconds. That’s roughly how fast things happen. RC = (10k+4.7k) * 10uF = 0.147 seconds. Your LED should blink roughly 8-10 times/sec.